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\title{\heiti\zihao{2} 复变函数-第4章}
\author{20373963-樊若宸}
\date{\today}

\begin{document}
\maketitle
\section{判断下列级数的敛散性}
(1)$\sum\limits_{n=1}^{\infty}\left(\dfrac{n+i}{n}\right)^{10}$

\textbf{解:}
令$u_n=\left(\dfrac{n+i}{n}\right)^{10}$,有$\lim\limits_{n\rightarrow \infty}u_n \neq 0$,从而级数发散.


(2)$\sum\limits_{n=2}^{\infty}\dfrac{i^n}{\ln n}$

\textbf{解:}
令$u_n=\dfrac{i^n}{\ln n}$,将其分为两个级数$u_{2k},u_{2k+1}$的和.显然两个级数分别是莱布尼茨交错级数,所以分别收敛.从而原级数收敛.

$|u_n|=\dfrac{1}{\ln n}$,级数$\sum\limits_{n=2}^{\infty}\dfrac{1}{\ln n}$发散.从而原级数条件收敛.


(3)$\sum\limits_{n=2}^{\infty}\dfrac{(3+5i)^n}{n!}$

\textbf{解:}
$$
\lim\limits_{n\rightarrow \infty}\sqrt[n]{\dfrac{1}{n!}} = \lim\limits_{n\rightarrow \infty}\dfrac{e}{n} = 0
$$
从而收敛半径$R=+\infty$,绝对收敛.

(4)$\sum\limits_{n=0}^{\infty}\dfrac{\sin in}{2^n}$

\textbf{解:}
令$u_n=\dfrac{\sin in}{2^n}=\dfrac{\mathrm{e}^{-n}-\mathrm{e}^{n}}{i2^{n+1}}$,有:
$$
\lim\limits_{n\rightarrow \infty}u_n = \lim\limits_{n\rightarrow \infty}\dfrac{\mathrm{e}^{-n}-\mathrm{e}^{n}}{i2^{n+1}} \neq 0
$$
从而级数发散.

\section{求下列幂级数的收敛半径和收敛圆圆心.}
(1) $\sum\limits_{n=1}^{\infty} \dfrac{n}{2^{n}} z^{n}$;

\textbf{解:}
$$
R = \lim\limits_{n\rightarrow \infty}\sqrt[n]{\dfrac{2^n}{n}} = 2 
$$
收敛圆圆心为$z=0$


(2) $\sum\limits_{n=1}^{\infty} \dfrac{z^{n}}{n^{n}}$;

\textbf{解:}
$$
R = \lim\limits_{n\rightarrow \infty}\sqrt[n]{n^n} = +\infty
$$
收敛圆圆心为$z=0$

(3) $\sum\limits_{n=0}^{\infty} \dfrac{(-1)^{n}}{n}(z-i)^{n}$

\textbf{解:}
$$
R = \lim\limits_{n\rightarrow \infty}\sqrt[n]{n} = 1
$$
收敛圆圆心为$z=i$

(4) $\sum\limits_{n=1}^{\infty} \dfrac{n !}{n^{n}} z^{n}$;

\textbf{解:}
$$
R = \lim\limits_{n\rightarrow \infty}\left|\dfrac{\dfrac{(n+1)!}{(n+1)^{n+1}}}{\dfrac{n!}{n^n}}\right|=\lim\limits_{n\rightarrow \infty}\left(1-\dfrac{1}{n+1}\right)^{n+1} = \dfrac{1}{\mathrm{e}}
$$
收敛圆圆心为$z=0$


(5) $\sum\limits_{n=1}^{\infty} \dfrac{\mathrm{e}^{n}}{n^{2}}(z-1)^{n}$;

\textbf{解:}
$$
R = \lim\limits_{n\rightarrow \infty}\sqrt[n]{\dfrac{n^2}{\mathrm{e}^n}} = \dfrac{1}{\mathrm{e}}
$$
收敛圆圆心为$z=1$

(6) $\sum\limits_{n=1}^{\infty} n^{\ln n} z^{n}$.

\textbf{解:}
$$
\begin{aligned}
    R &= \lim\limits_{n\rightarrow \infty} n^{\frac{\ln n}{n}}\\
    &=\lim\limits_{n\rightarrow \infty}\mathrm{e}^{\frac{(\ln n)^2}{n}}\\
    &=1
\end{aligned}
$$
收敛圆圆心为$z=0$
\section{求下列函数在$z=0$处得泰勒展式,并指出它们的收敛区域}
(1)$\dfrac{1}{(1+z)^2}$

(2)$\dfrac{\mathrm{e}^{z}}{1-z}$

\textbf{解:}\quad
$$
\begin{aligned}
    \dfrac{\mathrm{e}^{z}}{1-z}&=\left(\sum\limits_{k=0}^{\infty}\dfrac{z^{k}}{k!}\right)\cdot\left(\sum\limits_{k=0}^{\infty}z^k\right)\\
    &=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{\infty}\dfrac{1}{k!}\right)z^{n}
\end{aligned}
$$
收敛半径为$R=1$.当$|z|<1$时收敛.
(3)$\sin^2z$

(4)$\dfrac{1}{2}\left(\ln \dfrac{1}{1-z}\right)^2$

\textbf{解:}\quad
$$
\begin{aligned}
    \dfrac{1}{2}\left(\ln \dfrac{1}{1-z}\right)^2&=\dfrac{1}{2}\left(\ln(1-z)\right)^2\\
    &=\dfrac{1}{2}\left(\sum_{k=1}^{\infty}\dfrac{z^k}{k}\right)^2\\
    &=\sum_{k=2}^{\infty}\dfrac{1}{n}\left(\sum_{k=1}^{n-1}\dfrac{1}{k}\right)z^{n}
\end{aligned}
$$
收敛半径为$R=1$.$|z|<1$时收敛.
(5)$\arctan z$其中$\arctan z|_{z=0}=0$.

\section{写出下列函数的泰勒展式至含$z^4$项为止,并指出其收敛范围.}
(1)$\dfrac{\mathrm{e}^{z^2}}{\cos z}$

\textbf{解:}\quad
$$
\begin{aligned}
    \dfrac{\mathrm{e}^{z^2}}{\cos z}&=\left(\sum\limits_{k=0}^{\infty}\dfrac{z^{2k}}{k!}\right)\cdot\dfrac{1}{\cos z}\\
    &=\left(1+z^2+\dfrac{z^4}{2}+o\left(z^4\right)\right)\cdot\dfrac{1}{1-\left(\dfrac{z^2}{2}-\dfrac{z^4}{4!}+o(z^4)\right)}\\
    &=\left(1+z^2+\dfrac{z^4}{2}+o\left(z^4\right)\right)\cdot\left(1 + \left(\dfrac{z^2}{2}-\dfrac{z^4}{4!}\right)+\left(\dfrac{z^2}{2}-\dfrac{z^4}{4!}\right)^2+o(z^4)\right)\\
    &=1+\dfrac{3}{2}z^2+\dfrac{29}{24}z^4+o(z^4)
\end{aligned}
$$

收敛范围为$|z|<\dfrac{\pi}{2}$

(2)$\sin\dfrac{z}{1-z}$

\textbf{解:}\quad
$$
\begin{aligned}
    \sin \dfrac{z}{1-z}&=\sin\left(z+z^2+z^3+z^4+o(z^4)\right)\\
    &=\left(z+z^2+z^3+z^4\right)-\dfrac{\left(z+z^2+z^3+z^4\right)^3}{3!}+o(z^4)\\
    &=z+z^2+\dfrac{5}{6}z^3-\dfrac{1}{2}z^4+o(z^4)
\end{aligned}
$$

收敛范围为$|z|<1$.
\section{指出下列函数在零点$z=0$的阶数}
(1)$\mathrm{e}^{z^2}-1$

\textbf{解:}\quad
$\mathrm{e}^{z^2}-1=z^2+o(z)^2$,2阶.

(2)$3\cos z^4-3$

\textbf{解:}\quad
$3\cos z^4-3=-\dfrac{3}{2}z^8+o(z^8)$,为8阶.

\section{是否存在$z=0$处的解析函数$f(z)$,在$z_n=\dfrac{1}{n}$处取下列函数值?}
(1)0,1,0,1,0,1$\cdots$

函数在$z=0$不连续,不可能解析.


(2)$\dfrac{1}{2},\dfrac{1}{3},\dfrac{1}{4},\dfrac{1}{5},\dfrac{1}{6},\cdots$

存在,可取$f(z)=\dfrac{1}{1+\dfrac{1}{z}}$


(3)$\dfrac{1}{2},\dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{4},\dfrac{1}{6},\dfrac{1}{6},\cdots$

若存在,则不妨令$g(z)=\dfrac{1}{1+\dfrac{1}{z}},h(z)=z$,则取$z_{n}$的奇数项子列和偶数项子列代入$g,h$,则由于其分别和$f(z_{2k+1}),f(z_{2k})$各项相等,从而$f=g=h$,矛盾.

\end{document}